JEE Mains · Chemistry · STD 12 -1. Solution and colligative properties
\(\mathrm{CO}_{2}\) gas is bubbled through water during a soft drink manufacturing process at \(298\, \mathrm{~K}\). If \(\mathrm{CO}_{2}\) exerts a partial pressure of \(0.835\) bar then \(x \,\mathrm{~m} \,\mathrm{~mol}\) of \(\mathrm{CO}_{2}\) would dissolve in \(0.9\,\mathrm{~L}\) of water. The value of \(x\) is \(.....\) (Nearest integer) (Henry's law constant for \(\mathrm{CO}_{2}\) at \(298\, \mathrm{~K}\) is \(1.67 \times 10^{3}\) \(bar\))
- A \(50\)
- B \(25\)
- C \(55\)
- D \(35\)
Answer & Solution
Correct Answer
(B) \(25\)
Step-by-step Solution
Detailed explanation
From Henry's Law \(0.835=1.67 \times 10^{3} \times \frac{\mathrm{n}\left(\mathrm{CO}_{2}\right)}{\frac{0.9 \times 1000}{18}}\) \(\mathrm{n}\left(\mathrm{CO}_{2}\right)=0.025\) Millimoles of \(\mathrm{CO}_{2}=0.025 \times 1000=25\)
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